结构力学英文课件-chapter-3

Chapter3staticallydeterminatebeamsAbstractofthechaperTheobjectiveofthischapteristopresenttheanalysisofsupportreactionsandinternalforcethatmaydevelopinbeamsundertheactionofcoplanarsystemofexternalforcediagrams.Theanalysisofsinglespanbeamsisthefoundationoftheanalysisofmultispanbeam;thecontentpertinenttosinglespanbendingmomentdiagramsofstraightmembers,whichareplottedbyusingthemethodofsuperpositionsegmentbysegment.3.1theanalysisofsinglespanbeamsInengineeringpractice,singlespanbeamshaveavarietyofapplication.Theanalysispertainingtosinglespanbeamsisthefoundationofstaticallydeterminatemultispanbeams,framesandsoforth.Thecommonlyusedsinglespanbeamsarethefollowingthreekinds:(1)simplysupportedbeamsorsimplebeams[fig3.1],(2)cantileverbeamsand(3)overhangingbeamsThechapterwillreviewanddiscussthecalculationofinternalforcesandconstructionoftheirdiagramsofstaticallydeterminatesinglespanbeams.3.1.1internalforcesandtheirsignconventionForaplanarmember,itisgenerallysubjectedtoshearforcesandbendingmomentsaswellasaxialforcesundertheactionofexternalloads.Theinternalforceconponentorientedinthedirectionperpendiculartothecentroidalaxisofabeamatthesectionunderconsiderationisreferredtoastheaxialforce.Positiveaxialforcesmakethebeamintensilestatewhilenegativeincompressingstate.Theinternalforcecomponentorientedinthedirectionperpendiculartothecentroidalaxisofabeamatthesectionunderconsiderationiscalledtheshearforce.Theshearsareconsideredtobepositivewhentheytendtomaketheportionofthememberontheleftofthesectionrotateclockwiseanviceversa.Theinternalcouplewhichisthemomentaboutthesectionunderconsiderationofalltheinternalforcesistermedthebendingmoment.Thepositivebendingmomentisshowninfig.Thebendingmomentsareconsideredtobepositivewhentheytendtobendahorizontalbeamconcaveupward,causingcompressionintheupperfibersandtensioninthelowerfibersofthebeamatthesectionandviceversa.Themethodofcomputinginternalforces-methodofsectionsThemethodofsectionsisanelementarymethodtodeterminetheinternalforcecomponentsinamember.Themethodmeanstopassanimaginarysectionthroughthememberandcutthememberintotwopartsandthenconsiderthefreebodyoftheeithersideofthesection.wemayobtainthethreeinternalforcecomponentsatthesectionbyequilibriumequationsofthefreebody.Example3-1Figure3.3(a)showsasimplebeam,determinetheinternalforcesatsectionsC,D1andD2.1.befordeterminingtheinternalforcesthereactionscanbeobtainedbyconsideringtheequilibriumconditionsofthefreebodyoftheentirebeam2.TheinternalforcesatsectionCAnimaginarysectionispassedthroughsectionC,cuttingthebeamintotwoportions,ACandCB.TheportionACshowninFig.3.3(c),whichistotheleftsideofthesection,isusedheretocomputetheinternalforce.Byutilizingthethreeequilibriumconditions,theinternalforcescanbedetermined.

(tensioninthelowerfiber)Thenegativeindicatethattheirrealsenseareoppositetothoseshownonthefreebody,i.e.,axialforceisapressure.Theshearisreallynegative.Thesenseofispositivemeansthatitsactualdirectionisthesameasthatshownonthefreebody,thatiscausestensioninthelowerfibersofthebeamatsectionC.3.TheinternalforcesatsectionAnimaginarysectionispassedthroughsection,cuttingthebeamintoportions,.TheportionshowninFig.3.3(d),whichistotheleftsideofthesection,isusedheretocomputetheinternalforces,byutilizingthethreeequilibriumconditions,theinternalforcescanbedetermined.P374.TheinternalforcesatsectionAnimaginarysectionispassedthroughsection,cuttingthebeamintotwoportions,.theportionshowninFig.3.3(e),whichistotheleftsideofthesection,isusedheretocomputetheinternalforces.Byutilizingthethreeequilibriumconditions,theinternalforcescanbedetermined.Themeaningsoftheinternalforcesaresimilarastheforegoing.Theselectionofthefreebodycanbetheeithersideofthesectionunderconsiderationwhenusethemethodofsections.Forinstance,inaboveexamplewecanselecttheleftsideofthesectionunderconsiderationorrightsideofthesectionaswell.althougheitherofthetwosidesofthebeamcanbeemployedforcomputinginternalforces,weshouldselectthesidethatwillrequiretheleastamountofcomputationaleffort,suchasthesidethatdoesnothaveanyreactionsactingonitorthathastheleastnumberofexternalloadsandreactionsappliedtoit.basedonthisargument,selectsegment(Fig.3.3(f))andsegment(Fig.3.3(g))canfindinternalforcesmoreeasilyatsectionsrespectively.Itisrecommendedthatreaderscheckthem.Theprocedurefordetermininginternalforcesataspecifiedlocationonabeamcanbesummarizedasfollows;(1)passanimaginarysectionperpendiculartothecentroidalaxisofthebeamatthesectionwheretheinternalforcesaredesired,therebycuttingthebeamintotwoportions.Thenselectanyoneofthetwoportionsasthefreebody.(2)drawthefree-bodydiagramuponwhichreactionsandappliedloadsanddesiredinternalforcesshouldbeactuallyimposed.(3)applytheequilibriumconditionsofthefreebodyanddeterminethedesiredthreeinternalforces.Obviously,inordertoestablishthecorrectequilibriumequationssoastoobtaincorrectinternalforces,thekeystepispresentingthefree-bodydiagramcorrectly.followingconsiderationhastotakewhenpresentfree-bodydiagrams(1)Thefreebodyshouldbeisolatedcompletelywithitsrestraintsandalltherestraintsshouldbesubstitutedbytheircorrespondingconstraintforces.(约束力)(2)theconstraintforcesshouldmeetthecharacteristicsofrestraints.Forexample,whileremoveonerollersupport,onehingedsupportandonefixedsupport,onesupportreaction,twosupportreactionsandthreesupportreactionsshouldbeimposedonthefreebodyrespectively;whilecutonelink,onehingeandonerigidjoint,anaxialforce,anaxialforceandashear,anaxialforceandashearandacoupleshouldbeexertedonthefreebodyrespectively.(3)alltheforcesactingonthefreebodyshouldbeindicatedactually.Donotleaveouttheforcesdirectlyexertingonitanddonotsuperimposetheforcesnotdirectlyexertingonit.Theforcesexertingonafreebodycanbeclassifiedastwogroups.Oneistheexternallyappliedloadsandtheotherisinternalforcesexertedbytheotherportioncorrespondingtothefreebody.(4)Thepresentationofafreebodyshouldbeclearandaccurate,generally,giveforcesoughttoindicateaccordingtotheiractuallocationsandmagnitudesanddirections,whereasunknownforcesshouldbeindicatedattheirlocationsintheirpositivedirections.

Basedonthisregulation,ifconsequent(结果)unknownsarepositivetheyareactuallypositive;ifsomeoftheunknownsarenegativetheyareactuallynegative,theirdirectionindicationarejustintheoppositedirection.Therefore,youcanavoidtheambiguitycausedbydirectionsofunknownforces.Abovementionedprocedureofmethodofsectionsisanormalprogresstocalculateinternalforces.Itscounterpartmethod,whichdoesnotneedfree-bodydiagramandmaybedevelopedbythedefinitionofinternalforces,canbestatedasfollowing:Axialforce:theinternalaxialforceonanysectionofabeamisequalinmagnitudebutoppositeindirectiontothealgebraicsum(resultant)ofthecomponentsinthedirectionparalleltotheaxisofbeamofalltheexternalloadsandsupportreactionsactingoneithersideofthesectionunderconsideration.iftheportionofthebeamtoleftsideofthesectionisbeingusedforcomputingtheaxialforce,thentheexternalforcesactingtotheleftareconsideredpositive,whereastheexternalforceactingtotherightareconsideredtobenegative.Iftherightportionisbeingusedforanalysis,thentheexternalforcesactingtotherightareconsideredtobepositiveandviceversa.Shear:theshearonanysectionofabeamisequalinmagnitudebutoppositeindirectiontothealgebraicsumofthecomponentsinthedirectionperpendiculartotheaxisofthebeamofalltheexternalloadsandsupportreactionsactingoneithersideofthesectionunderconsideration.Iftheleftportionofthebeamisbeingusedforanalysis,thentheexternalforcesactingupwardareconsideredpositive,whereastheexternalforcesactingdownwardareconsideredtobenegative.iftherightportionhasbeenselectedforanalysis,thenthedownwardexternalforcesareconsideredpositiveandviceversa.Bendingmoment:thebendingmomentonanysectionofabeamisequalinmagnitudebutoppositeindirectiontothealgebraicsumofthemomentsaboutthecentroidofthecrosssectionunderconsiderationofalltheexternalloadsandsupportreactionsactingoneithersideofthesection.Iftheleftportionisbeingusedforanalysis,thentheclockwisemomentsareconsideredtobepositive,andthecounterclockwisemomentsareconsiderednegative.Iftherightportionhasbeenselectedforanalysis,thenthecounterclockwisemomentsareconsideredpositiveanviceversa.Example3-2p403.1.3relationshipsbetweenloads,shearsandbendingmomentsTheconstructionofinternalforcediagramscanbeconsiderablyexpeditedbyusingthebasicrelationshipsthatexistbetweentheloads,theshears,andthebendingmoments.(1)DifferentialrelationshipsToderivetheserelationships,considerabeamsubjectedtoanarbitaryloading,asshowninfig.3.4(a).Alltheexternalloadsshowninthisfigureareassumedtobeactingintheirpositivedirections.Asindicatedinthisfigure,theexternaldistributedandconcentratedloadsactingdownwardareconsideredpositive;theexternalcouplesactingclockwisearealsoconsideredtobepositiveandviceversa.Next,weconsidertheequilibriumofadifferentialelementoflengthdx,isolatedfromthebeambypassingimaginarysectionsatdistancexanx+dxfromtheoriginA,asshowninFig.3.4(a).Thefreebodydiagramoftheelementisshowninfig3.4(b),inwhichQandMrepresenttheshearandbendingmoment,respectively,actingontheleftfaceoftheelement,dQandDmdenotethechangesinshearandbendingmoment,respectively,overthedistancedx,Asthedistancedxisinfinitesimallysmall,thedistributedloadqactingontheelementcanbeconsideredtobeuniformofmagnitudeq(x).Inorderfortheelementtobeinequilibrium,theforcesandcouplesactingonitmustsatisfythetwoequationsofequilibrium,thethirdequilibriumequationisautomaticallysatisfied,sincenohorizontalforcesareactingontheelement,applyingtheequilibriumequationsthefollowingformulatecanbeobtained.①distributedloadsConsideringtheequilibriumofanarbitraryfreebodyofadifferentialelementoflengthdxisolatedfromthesegmentsubjectedtodistributedloadsofthebeam,wewrite(a)(b)(3-1)Combineeqs.(3-1)(a)and(b),weobtain(3-2)Abovedifferentialrelationshipsrepresentthefollowinggeometricmeanings:(a)canbeexpressedas:slopeofsheardiagramatapointequalstotheintensityofdistributedloadatthatpointbutoppositesign.(b)canbestatedas:slopeofbendingmomentdiagramatapointequalstotheshearatthatpoint.(3-2)canbepresentas:curvatureofbendingmomentdiagramatapointequalstotheintensityofdistributedloadatthatpointbutoppositesign.Basedonabovedifferentialrelationships,theinternalforcediagramslocatedinthesegmentsimposedbydistributedloadspossessthefollowinggeometriccharacteristics:a.atthesegmentimposedbyuniformlydistributedloads(q=constant)Qdiagramisalinearfunctionofx,thecurveofwhichisaninclinedstraightline.WhileMdiagramisaquadraticfunctionofx,themomentcurveisaparaboliccurve,whichisconcaveupwardforadownwardloadq②ConcentratedloadsTherelationshipsbetweentheloadsandshearsderivedthusfarthrougharenotvalidatthepointsofapplicationofconcentratedloads,becauseatsuchapointtheshearchangesabruptlybyanamountequaltothemagnitudeoftheconcentratedload.Toverifythisrelationship,weconsidertheequilibriumofadifferentialelementthatisisolatedfromthebeamoffig3.4(a)bypassingimaginarysectionsatinfinitesimaldistancestotheleftandtotherightofthepointofapplicationEoftheconcentratedloadp.thefree-bodydiagramofthiselementisshowninfig.3.4(c).Applyingtheequilibriumequationswewrite(3-3)Equation(3-3)(a)canbeexpressedas:themagnitudesofshearsatthetwosidesofthepointofapplicationofconcentratedloadareunequal,thesheardifferenceatthetwosidesisequaltothemagnitudeofP.therefore,thesheardiagramwillhaveabruptchangeatthepointofconcentratedloadpbyanamountequaltothemagnitudeoftheconcentratedloadp.(b)canbeexpressedas:thebendingmomentsatthetwosidesofthepointofapplicationofconcentratedloadareidentical.Notethatbecauseoftheabruptchangeinthesheardiagramatthatpoint.Thatis,therearecusps(anglepoints)onthemomentdiagramatthelocationswhereconcentratedloadsareappliedon.Whenconcentratedloadsaredownward,thecuspsaredownwardaswell.(ex)③couplesorconcentratedmomentsAlthoughtherelationshipsbetweentheloadsandshearsderivedthusfararevalidatthepointsofapplicationofcouplesorconcentratedmoments,therelationshipsbetweentheshearsandbendingmomentsasgivenbyeqs.(3-1)(b)and(3-3)(b)arenotvalidatsuchpoint,becauseatsuchapointthebendingmomentchangesabruptlybyanamountequaltothemagnitudeofthemomentorthecouple.Thederivethisrelationship,weconsidertheequilibriumofadifferentialelementthatisisolatedfromthebeamoffig.3.4(a)bypassingimaginarysectionsatinfinitesimaldistancetotheleftandtotherightofthepointFofapplicationofthecouplem.thefree-bodydiagramofthiselementisshowninfig.3.4(d).Applyingtheequilibriumequationswewrite(3-4)(a)canbestatedas:themagnitudesofshearsatthetwosidesofthepointofapplicationofconcentratedmomentmareequal;thuslythesheardiagramwillnotchangeatthetwosides.(b)canbestatedas:themagnitudesofbendingmomentsatthetwosidesofthepointofapplicationofconcentratedmomentmareequal,theirdifferenceatthetwosidesisequaltothemagnitudeofm,thereby,themomentdiagramwillhaveabruptchangeatthepointofapplicationofconcentratedmomentmbyanamountequaltothemagnitudeoftheconcentratedmomentm,whereastheslopeofthemomentdiagramwillnotchangesincetheconstantvalueofshearsatthetwosidesofthepoint,i.e.,theslopesatthetwosidesremainparallel.(2)integralrelationsipsTodeterminethechangeinshearandbendingmomentbetweenpointsAandBalongtheaxisofthemember(seefig.3.5),weintegrateEq.(3-1)fromAtoBtoobtainthefollowingintegralrelationsbetweenloadsandinternalforces[Eq.(3-5)].(3-5)Theseequationsimplythat:TheshearatendBisequaltothedifferencebetweentheshearatendAandtheresultantofloadqbetweenAandB,orthechangeinshearbetweensectionsAandBisequaltotheareaofthedistributedloaddiagrambetweensectionsAandB.

ThebendingmomentatendBisequaltothesumofthatatendAandtheareaofthesheardiagrambetweenAandB,orthechangeinbendingmomentbetweensectionsAandBisequaltotheareaofthesheardiagrambetweensectionsAandB.IfnotonlythedistributedloadsbutalsoconcentratedloadsareexertedonsegmentAB,Eq.(3-5)maybewrittenasWhererepresentsthealgebraicsumoftheverticalconcentratedloadsbetweenpointsAandB;isthesumoftheconcentratedmomentsexertedonsegmentAB.ThedownwardPandclockwisemareassumedtobepositiveintheequations.3.2ConstructionofbendingmomentdiagrambyprincipleofsuperpositionforstraightmembersTheprincipleofsuperpositionisextremelyconvenientinstructuralanalysis.Ifastructureissubjectedtoavarietyofloads,thedisplacementsinthestructurevarylinearlywiththeappliedloads,thatis,anyincrementindisplacementisproportionaltotheloadscausingit,andifalldeformationsofastructurearesmallenoughsothattheresultingdisplacementofthestructuredoesnotsignificantlyaffectthegeometryofthestructureandhencedonotchangetheinitiallyactingpropertyoftheloadsinthemembers.Undersuchconditions,reactions,internalforcesanddisplacementsduetotheloadscanbeobtainedbyutilizingtheprincipleofsuperposition.Theprinciplecanbeexpressedas,ifastructureislinearlyelastic,theforcesactingonthestructuremaybeseparatedordividedintosomeexpedientlyindividualformsandthenthestructuremaybeseparatelyanalyzedforeachindividualforms.Thefinalresultscanthenbeobtainedbyaddinguptheindividualresults.Nowwewilldiscusshowtheprincipleofsuperpositionbeusedtofacilitatetheconstructionofbendingmomentdiagramsforstraightmembers.3.2.1superpositionmethodofbendingmomentsforsimplebeamsFigure3.7(a)showsasimplysupportedbeamsubjectedtonotonlytheuniformlydistributedloadsinthespanbutalsotheendcouples.Byprincipleofsuperposition.Dividedtheloadsappliedonthebeamintotwogroups,oneisthetwoendcouples[fig.3.7(b)]andtheotherisuniformlydistributedloads[fig.3.7(c)]Ifthebeamundergoestheendcouplesonly,themomentdiagramissimplyastraightlineasshowninfig.3.7(b),itisdenotedbyM’diagram;andifthebeamissubjectedtotheuniformlydistributedloadsinthespanonly,thebendingmomentdiagramM(x)isasecond-orderparabolaandmaybeplottedasshowninfig.3.7(c)Thenthefinalbendingmomentdiagramofthebeammaybeobtainedbytheprincipleofsuperposition.Thatis,depictingtheM’diagramfirst,thentheordinatesofM(x)diagrambeingsuperimposedontheM’diagramwillcompletethefinalbendingmomentdiagramofthebeamasshowninfig3.7(d)Notethatitistheordinatesnotthefiguresofthetwodiagramswhicharesuperimposedtogether.TheordinateMatanyarbitrarysectionshowninfig3.7(d)shouldmeetthefollowingformula.M=M’+M(x)WheretheordinatesofM(x)areperpendiculartothebeamaxisABnotthedashedline.3.2.2superpositionmethodsegmentbysegmentNowextendthesuperpositionmethoddiscussedinabovesubsectiontotheconstructionofbendingmomentforanyarbitrary（任意的）segment（部分）ofastraightmember(连续截面).Figure.3.8(a)showsasimplysupportedbeamsubjectedtotheuniformlydistributedloadsonlyinsegmentAB.ThebendingmomentsMAatsectionAandMBatsectionBaredeterminedbythemethodofsections.NextwewilldiscusshowtoconstructthebendingmomentdiagramofsegmentABbyutilizingprincipleofsuperposition.TakesegmentABofthebeamshowninFig.3.8(a)asanexampleandcomparethefreebodydiagramofthesegment[fig3.8(b)]withthesimplysupportedbeaminfig.3.8(c).Wemayfindthatinthesetwocasesboththeloadqandtheendcouplesarethesame.Byemploying使用

theequilibriumequations,wealsofindYA=QAYB=QB.Since